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Computing Shortest Paths in the Plane with Removable Obstacles - - PowerPoint PPT Presentation
Computing Shortest Paths in the Plane with Removable Obstacles - - PowerPoint PPT Presentation
s t Computing Shortest Paths in the Plane with Removable Obstacles Pankaj K. Agarwal, Neeraj Kumar, Stavros Sintos and Subhash Suri Duke University and UC Santa Barabara Problem Description s t Input: h polygonal obstacles with n vertices,
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s t
Problem Description
Input: h polygonal obstacles with n vertices, source s and target t An obstacle Pi can be removed by paying cost ci
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s t
Problem Description
Input: h polygonal obstacles with n vertices, source s and target t An obstacle Pi can be removed by paying cost ci Assumption: Obstacles are mutually disjoint and convex
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s t
Problem Description
Input: h polygonal obstacles with n vertices, source s and target t Given a cost budget C, which obstacles should we remove An obstacle Pi can be removed by paying cost ci so that length of path from s to t is minimized? Assumption: Obstacles are mutually disjoint and convex
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Motivation
Classically, we deal with finding shortest path in a fixed network
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Motivation
Classically, we deal with finding shortest path in a fixed network But what if shortest paths are not good enough!
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Motivation
Classically, we deal with finding shortest path in a fixed network What if we are allowed to modify the environment What changes should we make?
Desire paths
But what if shortest paths are not good enough!
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Motivation
Classically, we deal with finding shortest path in a fixed network What if we are allowed to modify the environment What changes should we make? Applications But what if shortest paths are not good enough! Urban Planning Adding ‘shortcuts” to ease congestion
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Motivation
Classically, we deal with finding shortest path in a fixed network What if we are allowed to modify the environment What changes should we make? Applications Reconfiguring road networks – such as by adding flyovers But what if shortest paths are not good enough!
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Motivation
Classically, we deal with finding shortest path in a fixed network What if we are allowed to modify the environment What changes should we make? Applications Re-organize Warehouse Layout – Shorten frequent paths for robot But what if shortest paths are not good enough!
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Path Planning Under Uncertainity
Obstacle Pi is present in input with independent probability βi, not present with probability (1 − βi)
Probability of existence = βi
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Path Planning Under Uncertainity
Obstacle Pi is present in input with independent probability βi, not present with probability (1 − βi)
Probability of existence = βi
π1 π2 Path π1 has probability 1 Path π2 has probability (1 − βi)
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Path Planning Under Uncertainity
Obstacle Pi is present in input with independent probability βi, not present with probability (1 − βi)
Probability of existence = βi
π1 π2 Path π1 has probability 1 Path π2 has probability (1 − βi) What is the shortest s–t path that has probability at least β?
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Path Planning Under Uncertainity
Obstacle Pi is present in input with independent probability βi, not present with probability (1 − βi)
Probability of existence = βi
π1 π2 Path π1 has probability 1 Path π2 has probability (1 − βi) What is the shortest s–t path that has probability at least β? What is the most likely s–t path with length at most L?
(one that has highest probability)
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Path Planning Under Uncertainity
Obstacle Pi is present in input with independent probability βi, not present with probability (1 − βi)
Probability of existence = βi
π1 π2 Path π1 has probability 1 Path π2 has probability (1 − βi) What is the shortest s–t path that has probability at least β? What is the most likely s–t path with length at most L?
(one that has highest probability)
Can reduce to the earlier cost-based model by taking negative logarithms
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Related Work
Computes shortest s–t path in the plane that removes at most k obstacles O(k2n logn) algorithm using Continuous Dijkstra [HKS’17] Shortest Paths in the Plane with Obstacle Violations
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Related Work
Computes shortest s–t path in the plane that removes at most k obstacles O(k2n logn) algorithm using Continuous Dijkstra [HKS’17] Shortest Paths in the Plane with Obstacle Violations Cardinality Model Obstacles have unit cost of removal, budget = k
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Related Work
Computes shortest s–t path in the plane that removes at most k obstacles O(k2n logn) algorithm using Continuous Dijkstra [HKS’17] Shortest Paths in the Plane with Obstacle Violations Cardinality Model Obstacles have unit cost of removal, budget = k We study the more general cost-based model of obstacle removal
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Our Results
Problem of computing a path of length at most L and cost at most C is
NP-HARD even for vertical segment obstacles.
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Our Results
Problem of computing a path of length at most L and cost at most C is
NP-HARD even for vertical segment obstacles.
An FPTAS inspired from [HKS’17] running in worst case ˜
O(n2h/ϵ) time
Minimum Path length with cost at most (1 + ϵ)C
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Our Results
Problem of computing a path of length at most L and cost at most C is
NP-HARD even for vertical segment obstacles.
An FPTAS inspired from [HKS’17] running in worst case ˜
O(n2h/ϵ) time
A faster FPTAS running in worst case ˜
O(nh/ϵ2) time
Minimum Path length with cost at most (1 + ϵ)C Length (1 + ϵ) times optimal, cost at most (1 + ϵ)C
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Our Results
Problem of computing a path of length at most L and cost at most C is
NP-HARD even for vertical segment obstacles.
An FPTAS inspired from [HKS’17] running in worst case ˜
O(n2h/ϵ) time
A faster FPTAS running in worst case ˜
O(nh/ϵ2) time
Minimum Path length with cost at most (1 + ϵ)C Length (1 + ϵ) times optimal, cost at most (1 + ϵ)C
Data structures for fixed source approximate shortest path queries in
O(log2 n/ϵ) time, two-point queries in O(log2 n/ϵ2) time
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Our Results
Problem of computing a path of length at most L and cost at most C is
NP-HARD even for vertical segment obstacles.
An FPTAS inspired from [HKS’17] running in worst case ˜
O(n2h/ϵ) time
A faster FPTAS running in worst case ˜
O(nh/ϵ2) time
Minimum Path length with cost at most (1 + ϵ)C Length (1 + ϵ) times optimal, cost at most (1 + ϵ)C
Data structures for fixed source approximate shortest path queries in
O(log2 n/ϵ) time, two-point queries in O(log2 n/ϵ2) time
Simple reduction from Partition
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Our Results
Problem of computing a path of length at most L and cost at most C is
NP-HARD even for vertical segment obstacles.
An FPTAS inspired from [HKS’17] running in worst case ˜
O(n2h/ϵ) time
A faster FPTAS running in worst case ˜
O(nh/ϵ2) time
Minimum Path length with cost at most (1 + ϵ)C Length (1 + ϵ) times optimal, cost at most (1 + ϵ)C
Data structures for fixed source approximate shortest path queries in
O(log2 n/ϵ) time, two-point queries in O(log2 n/ϵ2) time
Simple reduction from Partition
Rest of the talk
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A Simple FPTAS
s t Observations
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A Simple FPTAS
s t Observations
The desired path π must ‘cross’ all the obstacles that were removed Since π has minimum length, it must only turn at obstacle vertices
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A Simple FPTAS
s t Observations
The desired path π must ‘cross’ all the obstacles that were removed Since π has minimum length, it must only turn at obstacle vertices
Define cost(π) = sum of costs of all obstacles crossed
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A Simple FPTAS
s t Observations
The desired path π must ‘cross’ all the obstacles that were removed Since π has minimum length, it must only turn at obstacle vertices
Define cost(π) = sum of costs of all obstacles crossed What is the shortest s–t path that has cost at most C?
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A Simple FPTAS
Scale all costs such that budget C = h, the number of obstacles
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A Simple FPTAS
Scale all costs such that budget C = h, the number of obstacles Model as a graph problem: – Nodes : s, t, obstacle vertices
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A Simple FPTAS
Scale all costs such that budget C = h, the number of obstacles Model as a graph problem: – Nodes : s, t, obstacle vertices – Edges : between any pair of vertices (u,v) such that cost(uv) is at most C
c1
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c2 c3
Cost of red edge is c1 + c2 + c3
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A Simple FPTAS
Scale all costs such that budget C = h, the number of obstacles Model as a graph problem: – Nodes : s, t, obstacle vertices – Edges : between any pair of vertices (u,v) such that cost(uv) is at most C An edge e has two parameters: Euclidean length ℓe and cost of the edge ce
c1
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c2 c3
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A Simple FPTAS
Scale all costs such that budget C = h, the number of obstacles Model as a graph problem: – Nodes : s, t, obstacle vertices – Edges : between any pair of vertices (u,v) such that cost(uv) is at most C An edge e has two parameters: Euclidean length ℓe and cost of the edge ce
c1
Mimics the notion of visibility graphs, we call it a viability graph
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c2 c3
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A Simple FPTAS
Scale all costs such that budget C = h, the number of obstacles Model as a graph problem: – Nodes : s, t, obstacle vertices – Edges : between any pair of vertices (u,v) such that cost(uv) is at most C An edge e has two parameters: Euclidean length ℓe and cost of the edge ce Find shortest s–t path in this graph such that the cost of the path is at most C
c1
Mimics the notion of visibility graphs, we call it a viability graph
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c2 c3
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A Simple FPTAS
Scale all costs such that budget C = h, the number of obstacles Model as a graph problem: – Nodes : s, t, obstacle vertices – Edges : between any pair of vertices (u,v) such that cost(uv) is at most C An edge e has two parameters: Euclidean length ℓe and cost of the edge ce Find shortest s–t path in this graph such that the cost of the path is at most C
c1
Mimics the notion of visibility graphs, we call it a viability graph
(1 + ϵ)-approximation using Dijkstra’s algorithm in ˜ O( n2h ϵ ) time by creating O( h ϵ ) copies of vertices and edges
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c2 c3
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A Faster FPTAS
Key idea is to construct a sparse Viability Graph
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A Faster FPTAS
Key idea is to construct a sparse Viability Graph Roadmap – trade a factor (1 + ϵ) in path length for efficiency
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A Faster FPTAS
Key idea is to construct a sparse Viability Graph Roadmap – Construct an O(n logn) size viability graph VG1 preserving L1 distances – trade a factor (1 + ϵ) in path length for efficiency
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A Faster FPTAS
Key idea is to construct a sparse Viability Graph Roadmap – Construct an O(n logn) size viability graph VG1 preserving L1 distances – Create O( 1
ϵ ) copies of VG1, one per direction and combine them to obtain VGϵ
– trade a factor (1 + ϵ) in path length for efficiency
Preserves pairwise distances within a factor of (1 + ϵ)
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A Faster FPTAS
Key idea is to construct a sparse Viability Graph Roadmap – Construct an O(n logn) size viability graph VG1 preserving L1 distances – Create O( 1
ϵ ) copies of VG1, one per direction and combine them to obtain VGϵ
– trade a factor (1 + ϵ) in path length for efficiency
Preserves pairwise distances within a factor of (1 + ϵ)
– Run the modified Dijkstra on O(n
ϵ logn) size viability graph VGϵ
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A Faster FPTAS
Key idea is to construct a sparse Viability Graph Roadmap – Construct an O(n logn) size viability graph VG1 preserving L1 distances – Create O( 1
ϵ ) copies of VG1, one per direction and combine them to obtain VGϵ
– trade a factor (1 + ϵ) in path length for efficiency
Preserves pairwise distances within a factor of (1 + ϵ)
– Run the modified Dijkstra on O(n
ϵ logn) size viability graph VGϵ
Primary challenge is to construct the graph VG1
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A Sparse Viability Graph
Inspiration: L1 Visibility Graphs : [Clarkson-Kapoor-Vaidya’87]
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A Sparse Viability Graph
Inspiration: L1 Visibility Graphs : [Clarkson-Kapoor-Vaidya’87] Main Idea: u
Do not need all O(n) edges adjacent to u
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A Sparse Viability Graph
Inspiration: L1 Visibility Graphs : [Clarkson-Kapoor-Vaidya’87] Main Idea: u v
Do not need all O(n) edges adjacent to u
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A Sparse Viability Graph
Inspiration: L1 Visibility Graphs : [Clarkson-Kapoor-Vaidya’87] Main Idea: u v
Do not need all O(n) edges adjacent to u Take projections on a ‘split line’ ℓ
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A Sparse Viability Graph
Inspiration: L1 Visibility Graphs : [Clarkson-Kapoor-Vaidya’87] Main Idea: u v
Do not need all O(n) edges adjacent to u Take projections on a ‘split line’ ℓ
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A Sparse Viability Graph
Inspiration: L1 Visibility Graphs : [Clarkson-Kapoor-Vaidya’87] Main Idea: u v
Do not need all O(n) edges adjacent to u Take projections on a ‘split line’ ℓ
u v
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Also need at most four projections on neighboring obstacles
A Sparse Viability Graph
Inspiration: L1 Visibility Graphs : [Clarkson-Kapoor-Vaidya’87] Main Idea: u v
Do not need all O(n) edges adjacent to u Take projections on a ‘split line’ ℓ
u v Total size: O(n logn)
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Also need at most four projections on neighboring obstacles
A Sparse Viability Graph
Inspiration: L1 Visibility Graphs : [Clarkson-Kapoor-Vaidya’87] Main Idea: u v
Do not need all O(n) edges adjacent to u Take projections on a ‘split line’ ℓ
u v Total size: O(n logn) We show how to construct O(n logn) size L1 Viability Graphs
edges can go through obstacles
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Algorithm Overview
Will reuse idea of taking split line projections for [CKV’87]
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Algorithm Overview
Will reuse idea of taking split line projections for [CKV’87]
can be naturally extended to edges of viability graph
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Algorithm Overview
Will reuse idea of taking split line projections for [CKV’87]
can be naturally extended to edges of viability graph However, taking projections on four neighboring obstacles is not sufficient
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Algorithm Overview
Will reuse idea of taking split line projections for [CKV’87]
can be naturally extended to edges of viability graph However, taking projections on four neighboring obstacles is not sufficient We circumvent this problem by adding “bypass vertices” Using this model we give an arguably simpler proof
- f correctness for the more general viability graphs
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Algorithm Overview
Will reuse idea of taking split line projections for [CKV’87]
can be naturally extended to edges of viability graph However, taking projections on four neighboring obstacles is not sufficient We circumvent this problem by adding “bypass vertices” Using this model we give an arguably simpler proof
- f correctness for the more general viability graphs
Next we describe an algorithm for constructing an L1 viability graph VG1
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Obstacles with Costs ⇒ Segments with Costs
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Obstacles with Costs ⇒ Segments with Costs
π1 π2 Obstacles are convex and disjoint ⇒ Shortest paths intersect zero or exactly two sides of an obstacle
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Obstacles with Costs ⇒ Segments with Costs
π1 π2 Obstacles are convex and disjoint ⇒ Shortest paths intersect zero or exactly two sides of an obstacle Assign cost ci/2 to all segments of an obstacle that has cost ci
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Obstacles with Costs ⇒ Segments with Costs
π1 π2 Obstacles are convex and disjoint ⇒ Shortest paths intersect zero or exactly two sides of an obstacle Assign cost ci/2 to all segments of an obstacle that has cost ci Allows us to reason about geometry of line segments
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Algorithm Description
v w
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Algorithm Description
v w ℓ
Find the split line ℓ that splits vertices into equal sized sets Vl and Vr
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Algorithm Description
v w ℓ v′ w′
Find the split line ℓ that splits vertices into equal sized sets Vl and Vr Let v′ be the projection of v on this split line ℓ
– Add vertex v′ and edge vv′ with cost c(vv′) – Cost of projection segment is the sum of
costs of all obstacle segments that it intersects
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Algorithm Description
v w ℓ w2 v′ w′ v1 w1 v2
Find the split line ℓ that splits vertices into equal sized sets Vl and Vr Let v′ be the projection of v on this split line ℓ
– Add vertex v′ and edge vv′ with cost c(vv′)
Find first segment s′ of positive slope that intersects ℓ
– Add two bypass vertices v1,v2 and edges vv′ and v1v2 – Bypass edges v1v2 have cost c(v1v2) = 0 – Cost of projection segment is the sum of
costs of all obstacle segments that it intersects
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Algorithm Description
v w ℓ w2 v′ w′ v1 w1 v2
Find the split line ℓ that splits vertices into equal sized sets Vl and Vr Let v′ be the projection of v on this split line ℓ
– Add vertex v′ and edge vv′ with cost c(vv′)
Find first segment s′ of positive slope that intersects ℓ
– Add two bypass vertices v1,v2 and edges vv′ and v1v2
Connect consecutive vertices on ℓ and obstacle boundary
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Algorithm Description
v w ℓ w2 v′ w′ ℓ v w v1 w1 v2
Find the split line ℓ that splits vertices into equal sized sets Vl and Vr Let v′ be the projection of v on this split line ℓ
– Add vertex v′ and edge vv′ with cost c(vv′)
Find first segment s′ of positive slope that intersects ℓ
– Add two bypass vertices v1,v2 and edges vv′ and v1v2
Connect consecutive vertices on ℓ and obstacle boundary Repeat for a horizontal split line. Recurse on sets Vl and Vr .
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Correctness
Suffices to show: for any pair of obstacle vertices p,q there exists a path πpq such that the L1 length ∥πpq∥1 ≤ ∥pq∥1 and cost c(πpq) ≤ c(pq) Proof Idea:
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Correctness
Suffices to show: for any pair of obstacle vertices p,q there exists a path πpq such that the L1 length ∥πpq∥1 ≤ ∥pq∥1 and cost c(πpq) ≤ c(pq) Proof Idea: p q
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Correctness
Suffices to show: for any pair of obstacle vertices p,q there exists a path πpq such that the L1 length ∥πpq∥1 ≤ ∥pq∥1 and cost c(πpq) ≤ c(pq) Proof Idea: p sy sx Define a region (clipped rectangle) Rpq q
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Correctness
Suffices to show: for any pair of obstacle vertices p,q there exists a path πpq such that the L1 length ∥πpq∥1 ≤ ∥pq∥1 and cost c(πpq) ≤ c(pq) Proof Idea: p sy sx Define a region (clipped rectangle) Rpq q Proof by induction on number of vertices in Rpq – If Rpq contains an obstacle vertex – If Rpq does not contains an obstacle vertex
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Correctness
Suffices to show: for any pair of obstacle vertices p,q there exists a path πpq such that the L1 length ∥πpq∥1 ≤ ∥pq∥1 and cost c(πpq) ≤ c(pq) Proof Idea: p sy sx Define a region (clipped rectangle) Rpq q Proof by induction on number of vertices in Rpq – If Rpq contains an obstacle vertex – If Rpq does not contains an obstacle vertex (Induction Step) (Base Case)
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Induction Step
p q lr r
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Induction Step
p q lr r We show how to find an intermediate vertex r such that
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Induction Step
p q lr r We show how to find an intermediate vertex r such that such that ∥pr∥1 + ∥rq∥1 ≤ ∥pq∥1 and cost c(pr) + c(rq) ≤ c(pq)
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Induction Step
p q lr r We show how to find an intermediate vertex r such that such that ∥pr∥1 + ∥rq∥1 ≤ ∥pq∥1 and cost c(pr) + c(rq) ≤ c(pq) Details are technical, relies heavily on disjointness of obstacle segments
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Base Case
ℓ B C Identify three types of edges A, B and C p A q
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Base Case
ℓ B C Identify three types of edges A, B and C Identify three types of edges A, B and C – Type B and C edges cannot co-exist p A q
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Base Case
ℓ B C Identify three types of edges A, B and C Identify three types of edges A, B and C – Type B and C edges cannot co-exist Type B edge + vertical split line ≡ Type C edge + horizontal split line p A q
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Base Case
ℓ B Identify three types of edges A, B and C Identify three types of edges A, B and C – Type B and C edges cannot co-exist Type B edge + vertical split line ≡ Type C edge + horizontal split line Exists a path πpq in VG1 using projection and bypass vertices p A q
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Base Case
ℓ B Identify three types of edges A, B and C Identify three types of edges A, B and C – Type B and C edges cannot co-exist Type B edge + vertical split line ≡ Type C edge + horizontal split line Exists a path πpq in VG1 using projection and bypass vertices Easy to verify that πpq has same L1 length and cost as the segment pq p A q
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A Faster FPTAS
Key idea is to construct a sparse Viability Graph Roadmap – Construct an O(n logn) size viability graph VG1 preserving L1 distances – Create 1/ϵ copies of VG1, one per direction and combine them to obtain VGϵ – trade a factor (1 + ϵ) in path length for efficiency
Preserves pairwise distances within a factor of (1 + ϵ)
– Run the slower FPTAS on O(n
ϵ logn) size viability graph VGϵ
Primary challenge is to construct the graph VG1
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A Faster FPTAS
Key idea is to construct a sparse Viability Graph Roadmap – Construct an O(n logn) size viability graph VG1 preserving L1 distances – Create 1/ϵ copies of VG1, one per direction and combine them to obtain VGϵ – trade a factor (1 + ϵ) in path length for efficiency
Preserves pairwise distances within a factor of (1 + ϵ)
– Run the slower FPTAS on O(n
ϵ logn) size viability graph VGϵ
Primary challenge is to construct the graph VG1
✓
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In Summary
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In Summary
Motivated by applications, we study the problem of computing shortest
path among removable obstacles
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In Summary
Motivated by applications, we study the problem of computing shortest
path among removable obstacles
We show that computing such a path exactly is NP-hard.
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In Summary
Motivated by applications, we study the problem of computing shortest
path among removable obstacles
We show that computing such a path exactly is NP-hard. A simple (1 + ϵ)-approximation follows using the idea of viability
graphs (an extension of visibility graphs)
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In Summary
Motivated by applications, we study the problem of computing shortest
path among removable obstacles
We show that computing such a path exactly is NP-hard. A simple (1 + ϵ)-approximation follows using the idea of viability
graphs (an extension of visibility graphs)
Obtain a faster approximation by constructing a sparse viability graph
SLIDE 87
In Summary
Motivated by applications, we study the problem of computing shortest
path among removable obstacles
We show that computing such a path exactly is NP-hard. A simple (1 + ϵ)-approximation follows using the idea of viability
graphs (an extension of visibility graphs)
Obtain a faster approximation by constructing a sparse viability graph Obtained data structures for approximate shortest path queries running
in polylogarithmic time
SLIDE 88
In Summary
Motivated by applications, we study the problem of computing shortest
path among removable obstacles
We show that computing such a path exactly is NP-hard. A simple (1 + ϵ)-approximation follows using the idea of viability
graphs (an extension of visibility graphs)
Obtain a faster approximation by constructing a sparse viability graph Obtained data structures for approximate shortest path queries running
in polylogarithmic time
Applied these results to solve a shortest path problem in a stochastic
model of obstacles
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In Summary
Motivated by applications, we study the problem of computing shortest
path among removable obstacles
We show that computing such a path exactly is NP-hard. A simple (1 + ϵ)-approximation follows using the idea of viability
graphs (an extension of visibility graphs)
Obtain a faster approximation by constructing a sparse viability graph Obtained data structures for approximate shortest path queries running
in polylogarithmic time
Applied these results to solve a shortest path problem in a stochastic
model of obstacles
Thanks!
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Backup : NP-Hardness
Simple reduction from Partition ai t Gi a1 an . . . . . . s A is a set of integers, ai corresponds to group Gi
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Backup : NP-Hardness
Simple reduction from Partition ai t Gi a1 an . . . . . . s A is a set of integers, ai corresponds to group Gi
Go through the middle obstacle by paying cost ai Go around it by adding a detour of length ai
Path must go through the channel
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Backup : NP-Hardness
Simple reduction from Partition ai t Gi a1 an . . . . . . s A is a set of integers, ai corresponds to group Gi
Go through the middle obstacle by paying cost ai Go around it by adding a detour of length ai
An s–t path with length at most L = 1
2 ∑ai and cost at most C = 1 2 ∑ai exists
Path must go through the channel